The probability that a missile hits a target successfully is $0.75$. In order to destroy the target completely, at least three successful hits are required. Then the minimum number of missiles that have to be fired so that the probability of completely destroying the target is NOT less than $0.95$, is ______

$p = P(\text{hit}) = 0.75 = \dfrac{3}{4}$ $\text{Destruction} \iff (\text{number of hits}) \ge 3$ $P(\text{target destroyed}) \ge 0.95$ $\text{To find: the minimum number of missiles } n$ $X = \text{number of hits in } n \text{ missiles}$ $X \sim B\!\left(n,\ \dfrac{3}{4}\right)$ $1.\ P(X \ge 3) = 1 - P(X \le 2)$ $2.\ P(X \le 2) \le 0.05$ $3.\ \text{test } n = 1, 2, 3, \dots$ $P(X \ge 3) = 1 - P(X \le 2) \ge 0.95$ $\Rightarrow P(X \le 2) \le 0.05$ $P(X = k) = \dbinom{n}{k}\left(\dfrac{3}{4}\right)^{k}\left(\dfrac{1}{4}\right)^{n-k}$ $P(X = 0) = \left(\dfrac{1}{4}\right)^{n}$ $P(X = 1) = n\cdot\dfrac{3}{4}\left(\dfrac{1}{4}\right)^{n-1} = 3n\left(\dfrac{1}{4}\right)^{n}$ $P(X = 2) = \dbinom{n}{2}\left(\dfrac{3}{4}\right)^{2}\left(\dfrac{1}{4}\right)^{n-2} = 9\dbinom{n}{2}\left(\dfrac{1}{4}\right)^{n}$ $P(X \le 2) = \left(\dfrac{1}{4}\right)^{n}\left[1 + 3n + 9\dbinom{n}{2}\right]$ $= \left(\dfrac{1}{4}\right)^{n}\left[1 + 3n + \dfrac{9n(n-1)}{2}\right]$ $n = 5:\ 1 + 3(5) + 9\dbinom{5}{2} = 1 + 15 + 90 = 106$ $P(X \le 2) = \dfrac{106}{4^{5}} = \dfrac{106}{1024} \approx 0.1035$ $0.1035 > 0.05 \Rightarrow P(X \ge 3) \approx 0.897 < 0.95\ \text{✗}$ $n = 6:\ 1 + 3(6) + 9\dbinom{6}{2} = 1 + 18 + 135 = 154$ $P(X \le 2) = \dfrac{154}{4^{6}} = \dfrac{154}{4096} \approx 0.0376$ $0.0376 \le 0.05 \Rightarrow P(X \ge 3) \approx 0.962 \ge 0.95\ \checkmark$ $P(X \ge 3) \text{ increases with } n$ $\Rightarrow \text{minimum } n = 6$ $n = 6:\ P(X \ge 3) = 1 - \dfrac{154}{4096} \approx 0.9624 \ge 0.95\ \checkmark$ $n = 5:\ P(X \ge 3) \approx 0.8965 < 0.95\ \text{✗}$ $\boxed{n = 6}$